从中值下凸证明下凸

2023-05-26 14:34 作者:~Sakuno酱 0人读过 | 我要投稿

$f(x)$ 在开区间 $I$ 上连续, 对任意 $a%2Cb%20%20%5Cin%20I$ 有 $f%5Cleft(%5Cfrac%7Ba%2Bb%7D%7B2%7D%5Cright)%5Cle%20%5Cfrac%7Bf%5Cleft(a%5Cright)%2Bf%5Cleft(b%5Cright)%7D%7B2%7D$

证明任取 $a%2Cb%20%20%5Cin%20I$ 对任意 $0%3Ct%3C1$ 有 $f(a%20%2B%20t(b%20-%20a))%20%5Cle%20f(a)%20%2B%20t%20(f(b)%20-%20f(a))$

第一步:证明引理

对任意 $1%5Cle%20n%2C0%3Cm%3C2%5E%7Bn%7D$

$f%20(a%20%2B%5Cfrac%7Bb%20-a%7D%7B2%5E%7Bn%7D%7D%5Ccdot%20m)%5Cle%20f%20(a)%2B%5Cfrac%7Bf%20(b)-f%20(a)%7D%7B2%5E%7Bn%7D%7D%5Ccdot%20m%20$

归纳法假设

$n%3D1$ 时显然成立

考虑 $n%2B1$

$f%20(a%20%2B%5Cfrac%7Bb%20-a%7D%7B2%5E%7Bn%20%2B1%7D%7D%5Ccdot%20m)%3Df%20(%5Cfrac%7Ba%20%2Ba%20%2B%5Cfrac%7Bb%20-a%7D%7B2%5E%7Bn%7D%7D%5Ccdot%20m%7D%7B2%7D)$ $%5Cle%20%5Cfrac%7Bf%5Cleft(a%5Cright)%2Bf%5Cleft(a%2B%5Cfrac%7Bb-a%7D%7B2%5E%7Bn%7D%7D%5Ccdot%20m%5Cright)%7D%7B2%7D$

$%3D%5Cfrac%7Bf%5Cleft(a%5Cright)%2Bf%5Cleft(a%5Cright)%2B%5Cfrac%7Bf%5Cleft(b%5Cright)-f%5Cleft(a%5Cright)%7D%7B2%5E%7Bn%7D%7D%5Ccdot%20m%7D%7B2%7D%3Df%5Cleft(a%5Cright)%2B%5Cfrac%7Bf%5Cleft(b%5Cright)-f%5Cleft(a%5Cright)%7D%7B2%5E%7Bn%2B1%7D%7D%5Ccdot%20m$

第二步: 构造数列 $q_n$ 去逼近 $t$ , 再利用连续函数和数列极限的性质

$q_%7Bn%7D%20%3D%20%5Cfrac%7B%5Cleft%5Bt%5Ccdot%202%5E%7Bn%7D%5Cright%5D%7D%7B2%5E%7Bn%7D%7D$ $%5Cleft%5Bx%5Cright%5D$ 代表取整函数显然 $%5Clim_%7Bn%5Cto%20%5Cinfty%7D%20q_n%20%3D%20t$

$f(a%20%2B%20(b%20-%20a)%5Ccdot%20t)$ $%3D%5Clim_%7Bn%5Cto%20%5Cinfty%7D%20f(a%20%2B%20(b%20-%20a)%5Ccdot%20q_n)%20$ $%5Cle%20%5Clim_%7Bn%5Cto%20%5Cinfty%7D%20%20f%5Cleft(a%5Cright)%2B%5Cleft(f%5Cleft(b%5Cright)-f%5Cleft(a%5Cright)%5Cright)%5Ccdot%20q_%7Bn%7D$ $%3Df%5Cleft(a%5Cright)%2Bt%5Ccdot%20%5Cleft(f%5Cleft(b%5Cright)-f%5Cleft(a%5Cright)%5Cright)$

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